Problem: Evaluate the definite integral. $\int^{-2}_{-1}\left(\dfrac{16-x^3}{x^3}\right)\,dx = $
Solution: First, use the power rule: $\begin{aligned}\int^{-2}_{-1}\left(\dfrac{16-x^3}{x^3}\right)\,dx~&=~\int^{-2}_{-1}\left(\dfrac{16}{x^3}-\dfrac{x^3}{x^3}\right)\,dx \\&=~\int^{-2}_{-1}\left(16{x^{-3}}-{1}\right)\,dx \\&=(-8x^{-2}-x)\Bigg|^{-2}_{{-1}}\end{aligned}$ Second, plug in the limits of integration: $[-8\cdot({-2})^{-2}-({-2})]-[-8\cdot({-1})^{-2}-({-1})] = 0+7 = 7$. The answer: $\int^{-2}_{-1}\left(\dfrac{16-x^3}{x^3}\right)\,dx= 7$